Problem: What is the value of $\dfrac{d}{dx}\sin\left(2x-\dfrac{\pi}{3}\right)$ at $x=\pi$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{\sqrt{3}}{2}$ (Choice B) B $\dfrac12$ (Choice C) C $1$ (Choice D) D $-\sqrt 3$
Solution: Let's start by finding the expression for $\dfrac{d}{dx}\sin\left(2x-\dfrac{\pi}{3}\right)$. Then, we can evaluate it at $x={\pi}$. $\sin\left(2x-\dfrac{\pi}{3}\right)$ is a trigonometric expression, but its argument isn't simply $x$. Therefore, it defines a composite trigonometric function. In other words, suppose $u(x)=2x-\dfrac{\pi}{3}$, then $\sin\left(2x-\dfrac{\pi}{3}\right)=\sin\Bigl(u(x)\Bigr)$. $\dfrac{d}{dx}\sin\left(2x-\dfrac{\pi}{3}\right)$ can be found using the following identity: $\dfrac{d}{dx}\left[\sin\Bigl(u(x)\Bigr)\right]={\cos\Bigl(u(x)\Bigr)\cdot u'(x)}$ [Why is this identity true?] Let's differentiate! $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\sin\left(2x-\dfrac{\pi}{3}\right) \\\\ &=\dfrac{d}{dx}\sin\Bigl(u(x)\Bigr)&&\gray{\text{Let }u(x)=2x-\dfrac{\pi}{3}} \\\\ &={\cos\Bigl(u(x)\Bigr)\cdot u'(x)} \\\\ &={\cos\Bigl(2x-\dfrac{\pi}3\Bigr)\cdot 2}&&\gray{\text{Substitute }u(x)\text{ back}} \end{aligned}$ Now let's evaluate $\dfrac{d}{dx}\sin\left(2x-\dfrac{\pi}{3}\right)$ at $x={\pi} $. $\begin{aligned} &\phantom{=}{\cos\Bigl(2({\pi})-\dfrac{\pi}3\Bigr)\cdot 2} \\\\ &=2\cos\left(\dfrac{5\pi}{3}\right) \\\\ &=2\cdot \dfrac12 \\\\ &=1 \end{aligned}$ In conclusion, the value of $\dfrac{d}{dx}\sin\left(2x-\dfrac{\pi}{3}\right)$ at $x=\pi$ is $1$.